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ICE Princess25 [194]

3 years ago

12

What is the reduction half-reaction for 2Mg + O2 --> 2MgO?

1 answer:

Anestetic [448]3 years ago

8 0

Answer:

Here every atom of oxygen takes in 2 electrons, which means that a molecule of oxygen will take in 4 electrons. On the other hand, the oxidation state of magnesium is going from 0 on the reactants' side to +2 on the products' side, which means that magnesium is being oxidize

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Rzqust [24]

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3 0

2 years ago

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Concerning the 10.0 mL of 0.50 M NaCl 100 mL of soltion:Does the dilution change the concentration

disa [49]

The answer is affirmative because <span>dilutions give you a weaker Molarity
Plase see this in the dilution formula that follows:

C1V1 = C2V2
(0.50M) (10.00ml) = (C2) (100ml)
C2 = 0.050 M
10X's weaker
</span>Hope this helps a lot

6 0

4 years ago

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Serga [27]

5 because A=v/t 20/4=5

8 0

3 years ago

A 1.525g sample of a compound between nitrogen and hydrogen contains 1.333 g of nitrogen. Calculate its empirical formula. The e

vredina [299]

Answer:

NH2.

Explanation:

The mass of hydrogen in the sample = 1.525 - 1.333 = 0.192g.

Dividing the 2 masses by the relative atomic mass of hydrogen and nitrogen:

H: 0.192 / 1.008 = 0.1905

N: 1.333 / 14.007 = 0.09517

The ratio of N to H = 0.09517 : 0.1905

= 1 : 2.

So the empirical formula is NH2.

5 0

3 years ago

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

7 0

3 years ago